\(3x^3=21x\)
Tim x
Giải các phương trình sau:
a)5x-6=6+2x
b)10-3x/2=6x+1/3
c)(3x-8)*(3x-8-(9+2x)) = 0
d)(x+3/x-3)-(x-3/x+3)=48/x^2-9
e) (3x-8)*(7-21x)-(9+2x)*(7-21x)=0
tìm x, biết:
(x + 3)(x2 - 3x + 9) = 7x3 + 21x
\(\left(x+3\right)\left(x^2-3x+9\right)=7x^3+21x\\ \Leftrightarrow x^3+27=7x^3+21x\\ \Leftrightarrow6x^3+21x-27=0\\ \Leftrightarrow\left(6x^3-6x^2\right)+\left(6x^2-6x\right)+\left(27x-27\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6x^2+6x+27\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x^2+6x+27=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\6\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{51}{2}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\6\left(x+\dfrac{1}{2}\right)^2+\dfrac{51}{2}=0\left(vô.lí\right)\end{matrix}\right.\)
Vậy \(x=1\)
\(\Leftrightarrow x^3+27-7x^3-21x=0\)
\(\Leftrightarrow-6x^3-21x+27=0\)
\(\Leftrightarrow-6x^3+6x-27x+27=0\)
\(\Leftrightarrow-6x\left(x-1\right)\left(x+1\right)-27\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x^2+6x+27\right)=0\)
hay x=1
tim GTNN cua bt
a) \(x^4+x^2+2\)
b)\(3x^2-21x+15\)
c)\(x^2-4xy+5y^2+10x-22y+28\)
a) A = x4 + x2 + 2
Do : x4 ≥ 0 ∀x
x2 ≥ 0 ∀x
⇒ x4 + x2 + 2 ≥ 2
⇒ AMin = 2 ⇔ x = 0
b) B = 3x2 - 21x + 15
B = 3( x2 - \(2\dfrac{7}{2}x+\dfrac{49}{4}\) ) + 15 - \(\dfrac{147}{4}\)
B = 3( x - \(\dfrac{7}{2}\))2 - \(\dfrac{87}{4}\)
Do : 3( x - \(\dfrac{7}{2}\))2 ≥ 0 ∀x
⇒ 3( x - \(\dfrac{7}{2}\))2 - \(\dfrac{87}{4}\) ≥ - \(\dfrac{87}{4}\)
⇒ BMin = - \(\dfrac{87}{4}\) ⇔ x = \(\dfrac{7}{2}\)
c) C = x2 - 4xy + 5y2 + 10x - 22y + 28
C = x2 - 4xy + 4y2 + 10x - 20y + 25 + y2 - 2y + 1 + 2
C = ( x - 2y)2 + 10( x - 2y) + 25 + ( y - 1)2 + 2
C = ( x - 2y + 5)2 + ( y - 1)2 + 2
Do : ( x - 2y + 5)2 ≥ 0 ∀xy
( y - 1)2 ≥ 0 ∀y
⇒ ( x - 2y + 5)2 + ( y - 1)2 + 2 ≥ 2
⇒ CMin = 2 ⇔ x = - 3 ; y = 1
tim x thuoc N
a,\(21x-1⋮11\)
b,\(5x+1⋮7\) va x la STN lon nhat co 3 chu so
c,\(3x-2⋮11\) va \(100< x< 2017\)
d,\(5⋮\left(x-3\right)\)
e,\(\left(3x+2\right)⋮\left(x+1\right)\)
e (3x+2)\(⋮\) (x+1)
vì (x+1)\(⋮\) (x+1)
=> (3x+3)\(⋮\) (x+1)
=> (3x+2)-(3x+3)\(⋮\) (x+1)
=>(3x+2-3x-3)\(⋮\) (x+1)
=> -1\(⋮\) (x+1)
=> (x+1)\(\in\) Ư(-1)={-1;1}
ta có bảng sau
x+1 | -1 | 1 |
x | -2 | 0 |
loại | thỏa mãn |
vậy x=0
Tìm x
(3x-3)(5-21x)+(7x+4)(6x-5)=45
a)A= 2x² -3x+1 với x=-3 b)B=⅐x2²y×21x³y² với x=-1,y=2
tim ucln cua 14x+3
tim ucln cua 21x+1
giải pT
(g)\(^{x^{2
}}\)-3x+2=0
i) x^4 +x^2 +6x -8=0
h) x^3-8x^2+21x-18=0
g: =>(x-1)(x-2)=0
=>x=1 hoặc x=2
i: \(\Leftrightarrow x^4-x^3+x^3-x^2+2x^2-2x+8x-8=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-x+4\right)=0\)
=>x=1 hoặc x=-2
g) x^2 - 3x + 2 = 0
<=> x^2 - 2x-x+2 =0
<=> x=1 hoặc x = 2
..tự kết luận
i)x^4 + x^2 + 6x - 8=0
<=> x^4 + 2x^3 - 2x^3 - 4x^2 + 5x^2 + 10x - 4x - 8 = 0
<=> x^3(x + 2) - 2x^2(x+2) + 5x(x+2) - 4(x+2) = 0
<=> (x^3 - 2x^2 +5x -4)(x+2)=0
<=> (x^3 - x^2 -x^2 +x + 4x - 4)(x+2) = 0
<=>(x^2(x-1) - x(x-1) + 4(x-1) )(x+2) = 0
<=> (x^2-x+4)(x-1)(x+2)=0
<=> x = 1 hoặc x +-2 hoặc x^2 - x+4=0
<=>x^2 - x+ 1/4 - 1/4 +4=0
<=>(x-1/2)^2 +15/4=0
<=>(x-1/2)^2=-15/4 (vô lí)
....tự kết luận
h)x^3 - 8x^2 + 21x - 18 = 0
<=> x^3 - 2x^2 - 6x^2 + 12x + 9x - 18 = 0
<=> x^2(x-2) -6x(x-2) + 9(x-2) =0
<=>(x-3)^2(x-2)=0
<=> x=3 hoặc x =2
...tự kết luận
câu 6: (x-5)(x-4)=10-2x
câu 7:(x ²+1)(x-2)+2x=4
câu 8:(x+3)(x ²-3x+9)=7x ³+21x
Câu 7:
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)
6, (x-5).(x-4)=10-2x
(x-5).(x-4)+2(x-5)=0
(x-5)(x-2)=0
=>x=5, x=2
7, (x^2+1)(x-2)+2x=4
x^3-2x^2+x-2+2x=4
x^3-2x^2+3x-2-4=0
x^3-2x^2+3x-6=0
x^2(x-2)+3(x-2)=0
(x-2)(x^2+3)=0
th1: x=2
th2: x^2+3>0 với mọi x thuộc Z
8, ( đề cs sai hông , giải hong ra:>)